$\text { Find } a _ { \mathbb { N } } \text { at time } t = 2 \text { for the space curve } \mathbf { r } ( t ) = ( 5 t - 7 ) \mathbf { i } + t ^ { 2 } \mathbf { j } - 7 t \mathbf { k } \text {. Round your }$ answer to three decimal places.
A)0.970
B)0.465
C)0.930
D)1.814
E)1.060

Question

Quizplus · Accepted Answer

To find the normal acceleration $a_{\mathbb{N}}$ at time $t = 2$ for the given space curve $\mathbf{r}(t) = (5t - 7)\mathbf{i} + t^2\mathbf{j} - 7t\mathbf{k}$, we first need to calculate the velocity $\mathbf{v}(t)$ and acceleration $\mathbf{a}(t)$ vectors by differentiating $\mathbf{r}(t)$ with respect to $t$.1. Velocity $\mathbf{v}(t) = \mathbf{r}'(t) = 5\mathbf{i} + 2t\mathbf{j} - 7\mathbf{k}$.2. Acceleration $\mathbf{a}(t) = \mathbf{v}'(t) = 0\mathbf{i} + 2\mathbf{j} + 0\mathbf{k}$.At $t = 2$, $\mathbf{v}(2) = 5\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ and $\mathbf{a}(2) = 2\mathbf{j}$.The normal component of the acceleration is given by:$$a_{\mathbb{N}} = \sqrt{|\mathbf{a}|^2 - (\mathbf{a} \cdot \mathbf{\hat{v}})^2}$$where $\mathbf{\hat{v}}$ is the unit vector in the direction of $\mathbf{v}$.First, find the magnitude of $\mathbf{v}(2)$:$$|\mathbf{v}(2)| = \sqrt{5^2 + 4^2 + (-7)^2} = \sqrt{25 + 16 + 49} = \sqrt{90}$$The unit vector $\mathbf{\hat{v}}(2)$ is:$$\mathbf{\hat{v}}(2) = \frac{\mathbf{v}(2)}{|\mathbf{v}(2)|} = \frac{5\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}}{\sqrt{90}}$$The dot product $\mathbf{a}(2) \cdot \mathbf{\hat{v}}(2)$ is:$$\frac{2\mathbf{j} \cdot (5\mathbf{i} + 4\mathbf{j} - 7\mathbf{k})}{\sqrt{90}} = \frac{8}{\sqrt{90}}$$Now, calculate $a_{\mathbb{N}}$:$$a_{\mathbb{N}} = \sqrt{|\mathbf{a}(2)|^2 - \left(\frac{8}{\sqrt{90}}\right)^2} = \sqrt{2^2 - \left(\frac{8}{\sqrt{90}}\right)^2}$$$$a_{\mathbb{N}} = \sqrt{4 - \frac{64}{90}} = \sqrt{\frac{360 - 64}{90}} = \sqrt{\frac{296}{90}} = \sqrt{\frac{148}{45}}$$$$a_{\mathbb{N}} \approx 1.814$$Therefore, the correct answer is D) 1.814.

Answer to Three Decimal Places